A garden is 24 m long and 14 m wide. There is a path 1 m wide outside the garden along its sides. If the path is to be constructed with square marble tiles 20 cm ÃÂÃÂÃÂÃÂÃÂÃÂ 20 cm, the number of tiles required to cover the path is :
(A) 200
(B) 1800
(C) 2000
(D) 2150
Solution:
Area of the path : $$\eqalign{ & = \left[ {\left( {26 \times 16} \right) - \left( {24 \times 14} \right)} \right]{m^2} \cr & = \left( {416 - 336} \right){m^2} \cr & = 80\,{m^2} \cr} $$ ∴ Number of tiles required to cover the path : $$\eqalign{ & = \frac{{{\text{Area of path}}}}{{{\text{Area of each tile}}}} \cr & = \left( {\frac{{80 \times 100 \times 100}}{{20 \times 20}}} \right) \cr & = 2000 \cr} $$
72.
A man is walking in a rectangular field whose perimeter is 6 km. If the area of the rectangular field be 2 sq. km, then what is the difference between the length and breadth of the rectangle ?
(A)
(B)
(C)
(D)
Solution:
Let the length and breadth of the field be l and b km respectively Then, 2 (l + b) = 6 ⇒ l + b = 3 And lb = 2 (l - b)2 = (l + b)2 - 4lb ⇒ (l - b)2 = 32 - 4 × 2 ⇒ (l - b)2 = 1 ⇒ (l - b) = 1 km
73.
The ratio of the area of a square to that of the square drawn on diagonal is :
Two equal circle are drawn in square in such a way that a side of the square forms diameter of each circle. If the remaining area of the square is 42 cm2, how much will the diameter of the circle measure ?
(A) 3.5 cm
(B) 4 cm
(C) 14 cm
(D) 7.5 cm
Solution:
Let length of each side of square = 2π According to the question, $$\eqalign{ & \frac{{\pi {r^2}}}{2} + \frac{{\pi {r^2}}}{2} + 42 = {\text{Area of square}} \cr & \Rightarrow \pi {r^2} + 42 = 4{r^2} \cr & \Rightarrow 4{r^2} - \pi {r^2} = 42 \cr & \Rightarrow {r^2}\left( {4 - \frac{{22}}{7}} \right) = 42 \cr & \Rightarrow {r^2}\left( {\frac{{28 - 22}}{7}} \right) = 42 \cr & \Rightarrow \frac{{6{r^2}}}{7} = 42 \cr & \Rightarrow {r^2} = \frac{{42 \times 7}}{6} \cr & \Rightarrow {r^2} = 7 \times 7 \cr & \Rightarrow r = 7 \cr & \therefore 2r = 14\,cm \cr} $$
75.
Three plots having areas 110, 130 and 190 square metres are to be subdivided into flower beds of equal size. If the breadth of a bed is 2 metres, the maximum length of a bed van be :
(A) 5 m
(B) 11 m
(C) 13 m
(D) 19 m
Solution:
Maximum possible size of a flower bed : = (H.C.F. of 110, 130, 190) sq. m = 10 sq. m ∴ Maximum possible length = $$\left( {\frac{{10}}{2}} \right)$$ m = 5 m
76.
The length of a rectangle is increased by 60%. By what percent would the width have to be decreased so as to maintain the same area ?
(A)
(B)
(C)
(D)
Solution:
Let original length = x and original breadth = y Then, original area = xy New length : $$\eqalign{ & = \frac{{160x}}{{100}} \cr & = \frac{{8x}}{5} \cr} $$ Let the new breadth = z Then, $$\eqalign{ & \Rightarrow \frac{{8x}}{5} \times z = xy \cr & \Rightarrow z = \frac{{5y}}{8} \cr} $$ ∴ Decrease in breadth : $$\eqalign{ & = \left( {\frac{{3y}}{8} \times \frac{1}{y} \times 100} \right)\% \cr & = 37\frac{1}{2}\% \cr} $$
77.
The diagonal of a rectangle is cm and its area is 20 sq. cm. The perimeter of the rectangle must be :
The circumference of a circle is equal to the side of a square whose area measures 407044 sq.cm. What is the area of the circle ?
(A) 22583.2 sq.cm
(B) 32378.5 sq.cm
(C) 39483.4 sq.cm
(D) 41263.5 sq.cm
Solution:
Area of the square = 407044 cm2 Side of the square = $$\sqrt {407044} $$ cm = 638 cm Circumference of circle = 638 cm Let the radius of the circle be R cm Then, $$\eqalign{ & 2\pi R = 638 \cr & \Rightarrow R = \frac{{638 \times 7}}{{2 \times 22}} \cr & \Rightarrow R = 101.5\,cm \cr} $$ ∴ Area of the circle : $$\eqalign{ & = \left( {\frac{{22}}{7} \times 101.5 \times 101.5} \right)c{m^2} \cr & = 32378.5\,\,c{m^2} \cr} $$
79.
A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?
(A) 34
(B) 40
(C) 68
(D) 88
Solution:
We have: l = 20 ft and lb = 680 sq. ft. So, b = 34 ft. ∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
80.
The sides of a triangle are in ratio of ÃÂÃÂ . If the perimeter is 52 cm, then the length of the smallest side is :
(A) 9 cm
(B) 10 cm
(C) 11 cm
(D) 12 cm
Solution:
Ratio of sides = $$\frac{1}{2}:\frac{1}{3}:\frac{1}{4}$$ = 6 : 4 : 3 Perimeter = 52 cm So, sides are : $$\eqalign{ & \left( {52 \times \frac{6}{{13}}} \right)cm \cr & \left( {52 \times \frac{4}{{13}}} \right)cm\,\& \cr & \left( {52 \times \frac{3}{{13}}} \right)cm \cr} $$ So, a = 24 cm, b = 16 cm, and c = 12 cm ∴ Length of smallest side = 12 cm